Facilitated Diffusion
Help Questions
AP Biology › Facilitated Diffusion
Cells are exposed to a fluorescent dye that is polar and cannot cross lipid bilayers unaided. Fluorescence inside cells increases only when a specific membrane channel is present and open. The increase occurs without detectable ATP consumption and slows as internal dye concentration rises. Which explanation best accounts for dye entry into the cells?
The dye enters by facilitated diffusion through an open channel down its concentration gradient.
The dye enters by active transport because channels hydrolyze ATP to move solutes.
The dye enters by moving from low to high concentration due to channel selectivity.
The dye enters by simple diffusion because polar molecules cross membranes rapidly.
The dye enters by phagocytosis because small solutes require vesicle uptake.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is A, as the polar dye enters via facilitated diffusion down its gradient through the open channel without ATP, slowing as internal concentration rises toward equilibrium. The channel provides a selective pathway. No energy is consumed. A tempting distractor is B, which wrongly attributes ATP hydrolysis to channels, mixing with active transport. For polar solutes, verify gradient-driven entry via proteins without energy to identify facilitated diffusion.
A researcher studies transport of polar solute R across a membrane. With a functional transporter present, R enters cells rapidly when outside concentration exceeds inside concentration. When the transporter gene is deleted, R entry becomes negligible despite the same gradient, and ATP inhibitors have no effect in either strain. Which explanation best accounts for these results?
R enters by endocytosis, and deleting the gene prevents vesicle fusion with the membrane.
R requires a membrane protein to cross; it moves down its gradient by facilitated diffusion.
R moves from low to high concentration, and deleting the gene removes the energy source.
R is imported by an ATP-driven pump, and inhibitors fail because cells store ATP.
R crosses by simple diffusion, and deleting the gene increases membrane rigidity.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is A, as polar R requires the transporter for facilitated diffusion down its gradient, negligible without it, and ATP inhibitors have no effect, confirming passivity. Deletion removes the protein pathway. Gradients drive entry. A tempting distractor is C, which wrongly suggests ATP-driven pumping, misapplying energy needs. When gene deletion halts gradient-driven transport without ATP effects, recognize facilitated diffusion dependency on proteins.
A membrane carrier transports solute Q. When extracellular Q is 60 mM and intracellular Q is 6 mM, net influx occurs. Adding a molecule structurally similar to Q reduces influx, but ATP levels remain constant and no ATP is consumed by the carrier. Which explanation best accounts for reduced influx after adding the similar molecule?
The similar molecule competitively inhibits binding to the carrier used for facilitated diffusion.
The similar molecule increases endocytosis rate, diluting Q influx through the carrier.
The similar molecule reverses the gradient, forcing Q to move into cells by osmosis.
The similar molecule blocks ATP production, preventing active transport of Q into cells.
The similar molecule dissolves the membrane, eliminating the concentration gradient for Q.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is A, as the similar molecule competitively inhibits Q binding to the carrier, reducing facilitated diffusion down the gradient from 60 mM to 6 mM without ATP consumption. Inhibition blocks sites for passive transport. Constant ATP confirms no energy role. A tempting distractor is B, which incorrectly links to ATP blocking, based on active transport misconception. Use competitive inhibition and gradient persistence without energy to confirm facilitated diffusion.
A cell expresses a carrier for solute W. Outside W is 18 mM and inside W is 2 mM. When temperature is lowered, W uptake rate decreases, but ATP inhibitors still do not affect uptake. Which explanation best accounts for the temperature dependence of W uptake?
Endocytosis slows at low temperature, eliminating the main pathway for W entry.
Simple diffusion through lipids slows because carriers block the bilayer at low temperature.
Transport slows because W must move from low to high concentration when temperature decreases.
Facilitated diffusion requires ATP, and low temperature prevents ATP synthesis needed for transport.
Carrier-mediated facilitated diffusion slows at low temperature because protein conformational changes occur more slowly.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is A, as carrier-mediated facilitated diffusion of W down its gradient from 18 mM to 2 mM slows at low temperature due to reduced protein conformational flexibility, unaffected by ATP inhibitors. Temperature affects protein dynamics in passive transport. Gradients remain the driver. A tempting distractor is B, which incorrectly ties to ATP synthesis, confusing with active processes. When temperature impacts rate without energy effects, consider protein involvement in facilitated diffusion.
Two cell types have different numbers of the same glucose transporter in their plasma membranes. Both are placed in 10 mM extracellular glucose with 1 mM intracellular glucose. Cell type 1 shows a faster initial glucose uptake rate than cell type 2, and neither shows a measurable change in ATP concentration during uptake. Which explanation best accounts for the difference in initial uptake rates?
Cell type 1 has more transport proteins, increasing facilitated diffusion capacity down the glucose gradient.
Cell type 1 has a lower extracellular glucose concentration, increasing diffusion into the cell.
Cell type 1 has more vesicles, increasing glucose uptake by endocytosis of extracellular fluid.
Cell type 1 has more mitochondria, providing more ATP to power active glucose pumping.
Cell type 1 has a higher intracellular glucose concentration, increasing net influx by diffusion.
Explanation
This question illustrates how transporter protein density affects facilitated diffusion rates. Both cell types have the same concentration gradient (10 mM outside to 1 mM inside), but cell type 1 has more glucose transporters in its membrane, allowing faster facilitated diffusion down the gradient without ATP consumption. The lack of ATP change confirms this is passive transport, not active pumping. Choice D incorrectly suggests higher intracellular glucose would increase influx, but this would actually decrease the gradient and slow transport. When comparing facilitated diffusion rates, consider both the concentration gradient magnitude and the number of available transport proteins.
A cell is placed in a solution with 9 mM molecule U outside and 3 mM inside. The membrane has U carriers that bind and release U without ATP hydrolysis. If the number of carriers is reduced by half, the initial rate of U uptake decreases. Which explanation best accounts for the reduced uptake rate?
Fewer carriers cause U to move from low to high concentration, reducing net influx at the start.
Fewer carriers lower ATP production, preventing ATP-dependent pumps from importing U into the cell.
Fewer carriers increase exocytosis of U, which directly blocks U entry through any remaining carriers.
Fewer carriers increase membrane thickness, stopping simple diffusion of U through the bilayer.
Fewer carriers lower the capacity for facilitated diffusion of U down its concentration gradient without ATP.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is A, as facilitated diffusion relies on carrier proteins to transport molecule U down its concentration gradient from 9 mM outside to 3 mM inside without requiring ATP. Reducing the number of carriers by half decreases the overall transport capacity, leading to a slower initial uptake rate because fewer proteins are available to bind and shuttle U across the membrane. The process is passive and driven solely by the concentration gradient, with carriers facilitating the movement without energy input. A tempting distractor is B, which incorrectly assumes ATP is involved, reflecting the misconception that all protein-mediated transport requires energy like active transport. To distinguish facilitated diffusion from other transport types, always check if movement is down the gradient and ATP-independent.
A neuron’s membrane contains ligand-gated ion channels permeable to Na+. Extracellular Na+ is 145 mM and intracellular Na+ is 12 mM. When a neurotransmitter binds, Na+ influx increases rapidly even when ATP synthesis is blocked, and the influx stops if the channel protein is chemically cross-linked closed. Which explanation best accounts for the Na+ influx observed after neurotransmitter binding?
Na+ enters by cotransport with glucose, requiring a proton gradient as energy input.
Na+ moves from low to high concentration because channels reverse diffusion direction.
Na+ is transported by a pump that requires ATP hydrolysis to open the channel.
Na+ moves down its concentration gradient through an opened channel protein without ATP use.
Na+ crosses the membrane by dissolving in phospholipids and diffusing directly.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is A, as Na+ moves down its concentration gradient from 145 mM outside to 12 mM inside through an opened ligand-gated channel without ATP, even when ATP synthesis is blocked. The neurotransmitter binding opens the channel, allowing passive influx, and chemical cross-linking closes it, stopping movement. This highlights how channels provide a pathway for ions to diffuse based on gradients without energy. A tempting distractor is B, which incorrectly implies ATP is needed for channel opening, confusing facilitated diffusion with active transport. To identify facilitated diffusion, confirm if transport is gradient-driven and unaffected by ATP inhibition.
A cell is placed in a solution containing 20 mM glucose while the cytosol contains 2 mM glucose. The plasma membrane includes a specific glucose transporter protein. When ATP production is inhibited, glucose still enters the cell until intracellular and extracellular concentrations become similar, and uptake stops when the transporter is blocked by a competitive inhibitor. Which explanation best accounts for glucose movement into the cell under these conditions?
A glucose pump hydrolyzes ATP to move glucose into the cell against its gradient.
Glucose moves down its concentration gradient through a membrane transporter without ATP input.
Glucose enters by endocytosis, forming vesicles that bypass the membrane proteins.
Glucose diffuses through the lipid bilayer because it is nonpolar and small.
Glucose exits the cell through the transporter because transporters only export solutes.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is C, as glucose moves down its concentration gradient from 20 mM outside to 2 mM inside through a specific transporter protein without ATP, continuing until equilibrium even when ATP is inhibited. The transporter facilitates passive diffusion, and blocking it with a competitive inhibitor stops uptake, confirming protein involvement. This process equalizes concentrations without energy input, as gradients drive the net movement. A tempting distractor is B, which wrongly suggests ATP-dependent pumping against the gradient, stemming from the misconception that all solute transport is active. When evaluating transport mechanisms, assess if net movement aligns with the concentration gradient and persists without ATP.
Red blood cells contain aquaporin proteins in the plasma membrane. When cells are placed in a hypotonic solution, water enters quickly; when aquaporins are blocked, water entry is much slower. No ATP is consumed during either condition. Which explanation best accounts for the faster water movement when aquaporins are present?
Aquaporins increase membrane surface area by forming vesicles for water import.
Aquaporins provide a hydrophilic pathway for water to move down its gradient.
Aquaporins hydrolyze ATP to pump water into the cell against its gradient.
Aquaporins bind water and carry it from low to high concentration by diffusion reversal.
Aquaporins convert water into ions that diffuse through the membrane.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is B, as aquaporins form hydrophilic channels that allow water to move down its osmotic gradient into the cell in hypotonic solutions without ATP consumption. Blocking aquaporins slows water entry, showing their role in facilitating faster passive movement across the otherwise impermeable bilayer. The absence of ATP use confirms the process is passive and gradient-driven. A tempting distractor is A, which falsely claims aquaporins use ATP for pumping, based on the misconception that all membrane proteins require energy. For water transport questions, evaluate if movement follows osmotic gradients without energy input to recognize facilitated diffusion.
A bacterial cell membrane contains a channel protein selective for glycerol. When extracellular glycerol is 50 mM and intracellular glycerol is 5 mM, glycerol enters rapidly. When extracellular glycerol is reduced to 5 mM, net glycerol movement stops even though the channel remains present. ATP levels are unchanged across treatments. Which explanation best accounts for the change in net glycerol movement?
Net glycerol movement depends on the concentration gradient driving diffusion through the channel.
The channel changes glycerol into lipids, preventing movement when concentrations equalize.
The channel pumps glycerol out of the cell whenever extracellular glycerol decreases.
Glycerol must be endocytosed, and vesicle formation stops at lower solute concentrations.
The channel requires ATP, so glycerol entry stops when ATP becomes limiting at 5 mM.
Explanation
This question tests understanding of facilitated diffusion. The correct answer is B, as net glycerol movement through the channel is driven by the concentration gradient, entering rapidly from 50 mM to 5 mM but stopping when concentrations equalize at 5 mM without ATP changes. The channel allows passive diffusion down the gradient, and equilibrium halts net flux. Unchanged ATP levels emphasize the energy-independent nature. A tempting distractor is A, which wrongly attributes stopping to ATP limitation, confusing it with active transport. To analyze transport cessation, check if equilibrium eliminates the driving gradient in passive processes.